Radical, What is Radical

Which positive number, multiplied by itself, gives 9? It is 3, because ${3\cdot 3=3^2=9}.$ We say that 3 is the square root of 9 and write ${\sqrt9=3}.$ Which number, taken as a factor three times, gives 27? Again 3, because ${3\cdot 3\cdot 3=3^3=27}.$ So 3 is the cube root of 27, and we write ${\sqrt[3]{27}=3}.$

The side of a square with area 9 is ${\sqrt9=3}.$ The edge of a cube with volume 27 is ${\sqrt[3]{27}=3}.$

Taking a root is the inverse of raising to a power: it undoes it. The power $3^2$ takes us from 3 to 9, and the root $\sqrt9$ brings us back from 9 to 3. The sign $\sqrt{\phantom{0}}$ is called the radical sign, and an expression written with it, such as $\sqrt9,$ $\sqrt[3]{27}$ or ${2\sqrt{x+1}},$ is called a radical. The word comes from the Latin radix, which means “root”.

The number $n$ is the index of the root, and the number $a$ under the radical sign is the radicand. For a square root the index 2 is not written: $\sqrt[2]{a}$ is simply $\sqrt a.$ The root with index 3 is the cube root; then come the fourth root, the fifth root and so on.

Examples

• ${\sqrt{25}=5},$ because ${5^2=25}.$
• ${\sqrt{\frac14}=\frac12},$ because ${\left(\frac12\right)^2=\frac14}.$
• ${\sqrt{0.09}=0.3},$ because ${0.3^2=0.09}.$
• ${\sqrt[3]{8}=2},$ because ${2^3=8}.$
• ${\sqrt[3]{-8}=-2},$ because ${(-2)^3=-8}.$
• ${\sqrt[4]{81}=3},$ because ${3^4=81}.$
• ${\sqrt[5]{32}=2},$ because ${2^5=32}.$
• ${\sqrt[n]{0}=0}$ and ${\sqrt[n]{1}=1}$ for every index $n,$ because ${0^n=0}$ and ${1^n=1}.$

On this page: Roots · Fractional exponents · Radical formulas · Simplifying · Adding and subtracting · Multiplying · Dividing · Rationalizing the denominator · Powers and roots of radicals · Comparing · Graphs · Practice

Square Roots, Cube Roots and n-th Roots

Perfect squares and perfect cubes

A whole number whose square root is also a whole number is called a perfect square: 1, 4, 9, 16, 25, … A number whose cube root is a whole number is a perfect cube: 1, 8, 27, 64, … It is worth learning the first few by heart, because we find roots by recognizing these numbers.

$x$	$x^2$	$x^3$	$x^4$
1	1	1	1
2	4	8	16
3	9	27	81
4	16	64	256
5	25	125	625
6	36	216	1296
7	49	343	2401
8	64	512	4096
9	81	729	6561
10	100	1000	10000

Read the table from right to left to find roots: ${\sqrt{49}=7},$ ${\sqrt[3]{216}=6},$ ${\sqrt[4]{625}=5}.$ The next perfect squares are ${11^2=121},$ ${12^2=144},$ ${13^2=169},$ ${14^2=196}$ and ${15^2=225}.$

Two square roots, one radical sign

The equation ${x^2=9}$ has two solutions, 3 and $-3,$ because ${3^2=9}$ and ${(-3)^2=9}.$ Both numbers are square roots of 9. But the symbol $\sqrt9$ stands for only one of them, the one that is not negative: ${\sqrt9=3}.$ It is called the principal (or arithmetic) square root. The negative root is written ${-\sqrt9=-3},$ and the two together are written ${\pm\sqrt9=\pm 3}.$

For example, ${x^2=7}$ gives ${x=\pm\sqrt7},$ and ${x^3=-125}$ gives ${x=\sqrt[3]{-125}=-5}.$ This is how the simplest quadratic and cubic equations are solved.

Even and odd index

Every root with an even index (an even root, for short) behaves like the square root, and every root with an odd index (an odd root) behaves like the cube root.

	$n$ is even: $\sqrt a,$ $\sqrt[4]{a},$ $\sqrt[6]{a},$ …	$n$ is odd: $\sqrt[3]{a},$ $\sqrt[5]{a},$ …
${a\gt 0}$	The root is positive: ${\sqrt[4]{16}=2}$	The root is positive: ${\sqrt[3]{8}=2}$
${a=0}$	${\sqrt[4]{0}=0}$	${\sqrt[3]{0}=0}$
${a\lt 0}$	No real root: $\sqrt[4]{-16}$ does not exist	The root is negative: ${\sqrt[3]{-8}=-2}$
The equation ${x^n=a}$	Two solutions $\pm\sqrt[n]{a}$ for ${a\gt 0},$ one (${x=0}$) for ${a=0},$ none for ${a\lt 0}$	Exactly one solution, $\sqrt[n]{a}$

Why is there no $\sqrt{-4}?$ A square is never negative: ${2\cdot 2=4}$ and ${(-2)\cdot(-2)=4},$ so no real number multiplied by itself gives $-4.$ (Square roots of negative numbers lead to imaginary numbers, a topic of more advanced algebra.) An odd root of a negative number does exist, and the minus sign simply comes out in front:

For example, ${\sqrt[3]{-27}=-\sqrt[3]{27}=-3}$ and ${\sqrt[5]{-32}=-\sqrt[5]{32}=-2}.$

Do not mix up $\sqrt{-a}$ and $-\sqrt a.$ The number ${-\sqrt{16}=-4}$ exists, while $\sqrt{-16}$ is not a real number.

For the same reason, an expression with an even root is defined only when its radicand is not negative: $\sqrt{x-3}$ is defined for ${x\ge 3},$ while $\sqrt[3]{x-3}$ is defined for every $x.$

A root of a negative number can show that a problem has no solution. Suppose we must split 14 into two parts whose product is 60. If one part is $x,$ the other is ${14-x},$ so ${x(14-x)=60},$ that is ${x^2-14x+60=0}.$ The quadratic formula would give ${x=7\pm\sqrt{-11}}.$ There is no real number $\sqrt{-11},$ so 14 cannot be split in this way. Indeed, the product is largest when the two parts are equal, and then it is only ${7\cdot 7=49}.$

Roots that are not whole numbers

Most roots are not whole numbers. Take $\sqrt2.$ Since ${1^2=1}$ and ${2^2=4},$ the root lies between 1 and 2. Let us try decimals:

${1.4^2=1.96}$ and ${1.5^2=2.25},$ so ${1.4\lt\sqrt2\lt 1.5}$

${1.41^2=1.9881}$ and ${1.42^2=2.0164},$ so ${1.41\lt\sqrt2\lt 1.42}$

We can go on like this forever: ${\sqrt2=1.41421356\ldots}$ The decimal never ends and never repeats. The number $\sqrt2$ is irrational: it cannot be written as a fraction whose numerator and denominator are whole numbers. An old name for such a root is a surd.

So $\sqrt4$ and $\sqrt[3]{27}$ are natural numbers, while $\sqrt2,$ $\sqrt3,$ $\sqrt5,$ $\sqrt6,$ $\sqrt[3]{2}$ and $\sqrt[3]{10}$ are irrational. A few values are used very often:

${\sqrt2\approx 1.414},$   ${\sqrt3\approx 1.732},$   ${\sqrt5\approx 2.236},$   ${\sqrt{10}\approx 3.162}$

An irrational root is still an exact number. In calculations we keep it as $\sqrt2$ and replace it with a decimal only at the very end, if at all.

To estimate a root without a calculator, find the nearest perfect squares below and above the radicand. For example, ${49\lt 50\lt 64},$ so ${7\lt\sqrt{50}\lt 8},$ and since 50 is very close to 49, the root is only a little more than 7. In fact ${\sqrt{50}\approx 7.07}.$

Why $\sqrt2$ cannot be written as a fraction

Suppose ${\sqrt2=\frac pq},$ where $p$ and $q$ are natural numbers and the fraction is in lowest terms. Squaring gives ${p^2=2q^2},$ so $p^2$ is even, and then $p$ itself is even: ${p=2k}.$ Now ${4k^2=2q^2},$ that is ${q^2=2k^2},$ so $q$ is even too. But then the fraction $\frac pq$ can be reduced by 2, and we said it was in lowest terms. The contradiction shows that no such fraction exists.

Roots as Powers: Fractional Exponents

A root can be written as a power. To see how, take a positive number $a$ and look at $a^6.$ Its square root is $a^3,$ because ${a^3\cdot a^3=a^6}$: the exponent 6 is divided by 2. Its cube root is $a^2,$ because ${a^2\cdot a^2\cdot a^2=a^6}$: the exponent 6 is divided by 3. Now do the same with ${a=a^1}.$ Its square root should have the exponent $\frac12.$ And indeed, if exponents are added in the usual way, then

${a^{\frac12}\cdot a^{\frac12}=a^{\frac12+\frac12}=a^1=a}$

So $a^{\frac12}$ multiplied by itself gives $a,$ exactly like $\sqrt a.$ In the same way, ${a^{\frac13}\cdot a^{\frac13}\cdot a^{\frac13}=a},$ like $\sqrt[3]{a}.$ Exponents of this kind are called fractional (or rational) exponents.

The second formula says that a power of a root equals the root of the power. For example, the fourth power of the cube root of $a$ is the same number as the cube root of $a^4.$

A negative fractional exponent means “one over”, as with integer exponents: ${a^{-\frac mn}=\frac{1}{a^{\frac mn}}}.$

Examples

• ${9^{\frac12}=\sqrt9=3}$
• ${8^{\frac13}=\sqrt[3]{8}=2}$
• ${8^{\frac23}=\left(\sqrt[3]{8}\right)^2=2^2=4}$
• ${16^{\frac34}=\left(\sqrt[4]{16}\right)^3=2^3=8}$
• ${27^{\frac43}=\left(\sqrt[3]{27}\right)^4=3^4=81}$
• ${4^{-\frac12}=\frac{1}{4^{\frac12}}=\frac{1}{\sqrt4}=\frac12}$
• ${100^{1.5}}={100^{\frac32}}={\left(\sqrt{100}\right)^3}={10^3}={1000}$
• ${32^{0.4}}={32^{\frac25}}={\left(\sqrt[5]{32}\right)^2}={2^2}={4}$

Take the root first and the power second: the numbers stay small. Computing $16^{\frac34}$ as ${\sqrt[4]{16^3}=\sqrt[4]{4096}}$ gives the same 8, but with much more work.

Equal fractions give equal powers: ${a^{\frac23}=a^{\frac46}=a^{\frac69}},$ that is ${\sqrt[3]{a^2}=\sqrt[6]{a^4}=\sqrt[9]{a^6}}.$ The exponent may also be written as a decimal: ${a^{0.5}=\sqrt a},$ ${a^{0.25}=\sqrt[4]{a}},$ ${a^{1.5}=\sqrt{a^3}}.$ This is how a calculator finds roots: $\sqrt[5]{7}$ is entered as $7^{0.2},$ and ${7^{0.2}\approx 1.476}.$

All the rules of exponents remain true for fractional exponents. Here ${a\gt 0}$ and ${b\gt 0},$ and $p$ and $q$ are any rational numbers:

Rule	Example
${a^p\cdot a^q=a^{p+q}}$	${a^{\frac12}\cdot a^{\frac13}=a^{\frac12+\frac13}=a^{\frac56}}$
${\frac{a^p}{a^q}=a^{p-q}}$	${\frac{a^{\frac12}}{a^{\frac16}}=a^{\frac12-\frac16}=a^{\frac13}}$
${\left(a^p\right)^q=a^{pq}}$	${\left(a^{\frac13}\right)^2=a^{\frac23}}$
${(ab)^p=a^pb^p}$	${(8a)^{\frac13}=8^{\frac13}a^{\frac13}=2a^{\frac13}}$
${\left(\frac ab\right)^p=\frac{a^p}{b^p}}$	${\left(\frac{a}{9}\right)^{\frac12}=\frac{a^{\frac12}}{3}}$

A few more examples:

• ${a^{\frac27}\cdot a^{\frac37}=a^{\frac57}}$
• ${y^3\cdot y^{\frac16}=y^{3+\frac16}=y^{\frac{19}{6}}}$
• ${a^3\cdot a^{\frac13}\cdot a^{\frac23}}={a^{3+\frac13+\frac23}}={a^4}$
• ${\left(a^{\frac23}x^{\frac34}\right)^2=a^{\frac43}x^{\frac32}}$

Only positive bases. A power with a fractional exponent is defined only for a positive base. The root ${\sqrt[3]{-8}=-2}$ is fine, but $(-8)^{\frac13}$ is left undefined. Otherwise the equal exponents $\frac13$ and $\frac26$ would give two different answers: $-2$ and ${\sqrt[6]{(-8)^2}=\sqrt[6]{64}=2}.$

Radical Formulas

The rules for working with radicals are the rules of exponents written with radical signs. In the table, $n,$ $m$ and $k$ are natural numbers (every index is at least 2), and $a$ and $b$ are not negative. In the quotient rule ${b\gt 0}.$

Rule	Formula	Example
Root of a product	${\sqrt[n]{ab}=\sqrt[n]{a}\cdot\sqrt[n]{b}}$	${\sqrt{4\cdot 25}=\sqrt4\cdot\sqrt{25}=10}$
Root of a quotient	${\sqrt[n]{\frac ab}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}}$	${\sqrt{\frac{9}{16}}=\frac{\sqrt9}{\sqrt{16}}=\frac34}$
Power of a root	${\left(\sqrt[n]{a}\right)^m=\sqrt[n]{a^m}}$	${\left(\sqrt[3]{8}\right)^2=\sqrt[3]{64}=4}$
Root of a root	${\sqrt[n]{\sqrt[m]{a}}=\sqrt[nm]{a}}$	${\sqrt{\sqrt[3]{64}}=\sqrt[6]{64}=2}$
Changing the index	${\sqrt[nk]{a^{mk}}=\sqrt[n]{a^m}}$	${\sqrt[6]{5^3}=\sqrt5}$
A root raised to its own index	${\left(\sqrt[n]{a}\right)^n=a}$	${\left(\sqrt7\right)^2=7}$
Root of an $n$-th power, $n$ odd	${\sqrt[n]{x^n}=x}$ for every real $x$	${\sqrt[3]{(-5)^3}=-5}$
Root of an $n$-th power, $n$ even	${\sqrt[n]{x^n}=|x|}$ for every real $x$	${\sqrt{(-5)^2}=\sqrt{25}=5}$

The last two rules are easy to mix up: ${\sqrt[3]{x^3}=x},$ but ${\sqrt[4]{x^4}=|x|},$ the absolute value of $x.$ The reason is that an even root is never negative: ${\sqrt[2n]{x}\ge 0}$ for every ${x\ge 0}$ and every natural number $n.$ For ${x=-5}$ we get ${\sqrt{(-5)^2}=\sqrt{25}=5},$ which is $|{-5}|$ and not $-5.$ Compare also ${\left(\sqrt x\right)^2=x},$ which makes sense only for ${x\ge 0},$ with ${\sqrt{x^2}=|x|},$ which is defined for every $x.$

Why the formulas are true

Written as powers with fractional exponents, the formulas turn into the rules of exponents (for ${a=0}$ both sides of each formula are 0).

Root of a product.

${\sqrt[n]{ab}}={(ab)^{\frac1n}}={a^{\frac1n}\cdot b^{\frac1n}}={\sqrt[n]{a}\cdot\sqrt[n]{b}}$

Root of a quotient.

${\sqrt[n]{\frac ab}=\left(\frac ab\right)^{\frac1n}=\frac{a^{\frac1n}}{b^{\frac1n}}=\frac{\sqrt[n]{a}}{\sqrt[n]{b}}}$

Power of a root.

${\left(\sqrt[n]{a}\right)^m}={\left(a^{\frac1n}\right)^m}={a^{\frac mn}}={\sqrt[n]{a^m}}$

Root of a root.

${\sqrt[n]{\sqrt[m]{a}}=\left(a^{\frac1m}\right)^{\frac1n}=a^{\frac{1}{mn}}=\sqrt[nm]{a}}$

Changing the index.

${\sqrt[nk]{a^{mk}}=a^{\frac{mk}{nk}}=a^{\frac mn}=\sqrt[n]{a^m}}$

We accepted the rules for fractional exponents without proof, so a complete proof has to start from the definition of a root. Here it is for the product rule. The number ${\sqrt[n]{a}\cdot\sqrt[n]{b}}$ is not negative, and its $n$-th power is ${\left(\sqrt[n]{a}\right)^n\cdot\left(\sqrt[n]{b}\right)^n=ab}.$ So it is the $n$-th root of $ab.$ The other formulas are checked in the same way. For example, ${\left(\left(\sqrt[n]{a}\right)^m\right)^n}={\left(\left(\sqrt[n]{a}\right)^n\right)^m}={a^m},$ so ${\left(\sqrt[n]{a}\right)^m}$ is the $n$-th root of $a^m.$

The product rule works for any number of factors: ${\sqrt[n]{abc}=\sqrt[n]{a}\cdot\sqrt[n]{b}\cdot\sqrt[n]{c}}.$

With an odd index, the product and quotient rules are true for negative numbers too: ${\sqrt[3]{(-8)\cdot 27}}={\sqrt[3]{-8}\cdot\sqrt[3]{27}}={-6}.$ With an even index they are not: ${\sqrt{(-4)\cdot(-9)}}={\sqrt{36}}={6},$ but $\sqrt{-4}$ and $\sqrt{-9}$ do not exist. If $a$ and $b$ are both negative, use ${\sqrt{ab}=\sqrt{-a}\cdot\sqrt{-b}}$: for example, ${\sqrt{(-4)\cdot(-9)}}={\sqrt4\cdot\sqrt9}={6}.$

There is no rule for a sum or a difference. In general ${\sqrt{a+b}\ne\sqrt a+\sqrt b}$ and ${\sqrt{a-b}\ne\sqrt a-\sqrt b}.$ For example, ${\sqrt{9+16}=\sqrt{25}=5},$ but ${\sqrt9+\sqrt{16}=3+4=7}.$ For positive $a$ and $b$ the root of the sum is always the smaller one: ${\sqrt{a+b}\lt\sqrt a+\sqrt b}.$

Letters under a radical. From here on, unless we say otherwise, letters stand for numbers that are not negative, and every expression under an even root is assumed to be not negative (positive in a denominator). Then ${\sqrt{a^2}=a}$ and no absolute value signs are needed.

Simplifying Radicals

A radical is in simplest form when

• the radicand has no factor greater than 1 that is a perfect $n$-th power (for a square root: no perfect square factor);
• there is no fraction under the radical sign and no radical in a denominator;
• the index is as small as possible.

Taking a factor out of the radical

Example. Simplify $\sqrt8.$ The largest perfect square that divides 8 is 4, so

${\sqrt8}={\sqrt{4\cdot 2}}={\sqrt4\cdot\sqrt2}={2\sqrt2}$

Here $2\sqrt2$ means ${2\cdot\sqrt2}.$ A number written in front of a radical is multiplied by it and is called its coefficient. When no coefficient is written, it is 1. Do not confuse a coefficient with an index: $3\sqrt2$ is 3 times $\sqrt2,$ while $\sqrt[3]{2}$ is the cube root of 2.

• ${\sqrt{18}=\sqrt{9\cdot 2}=3\sqrt2}$
• ${\sqrt{50}=\sqrt{25\cdot 2}=5\sqrt2}$
• ${\sqrt{12}=\sqrt{4\cdot 3}=2\sqrt3}$
• ${\sqrt{72}=\sqrt{36\cdot 2}=6\sqrt2}$
• ${\sqrt[3]{54}=\sqrt[3]{27\cdot 2}=3\sqrt[3]{2}}$
• ${\sqrt[4]{32}=\sqrt[4]{16\cdot 2}=2\sqrt[4]{2}}$

Look for the largest perfect square factor. For $\sqrt{72}$ the factor 4 also works and gives ${\sqrt{72}=2\sqrt{18}},$ but 18 still contains the perfect square 9, so one more step is needed: ${2\sqrt{18}=2\cdot 3\sqrt2=6\sqrt2}.$

When the radicand is large, factor it into primes. Every pair of equal factors gives one factor in front of a square root (every triple in front of a cube root):

${\sqrt{360}}={\sqrt{2^3\cdot 3^2\cdot 5}}={\sqrt{2^2\cdot 3^2\cdot 10}}={2\cdot 3\sqrt{10}}={6\sqrt{10}}$

Letters are treated in the same way. Divide the exponent by the index: the quotient is the exponent outside, and the remainder stays inside.

• ${\sqrt{a^2x}=a\sqrt x}$
• ${\sqrt{x^5}=\sqrt{x^4\cdot x}=x^2\sqrt x}$
• ${\sqrt{98a^2x}}={\sqrt{49a^2\cdot 2x}}={7a\sqrt{2x}}$
• ${\sqrt[3]{54a^6b}}={\sqrt[3]{27a^6\cdot 2b}}={3a^2\sqrt[3]{2b}}$
• ${\sqrt[3]{a^3+a^3b^2}}={\sqrt[3]{a^3\left(1+b^2\right)}}={a\sqrt[3]{1+b^2}}$

If the sign of $a$ is not known, an even root needs the absolute value: ${\sqrt{a^2x}=|a|\sqrt x}.$ Radicals such as $\sqrt{10},$ $\sqrt{15}$ and $\sqrt[3]{4}$ have no perfect power factor, so they are already in simplest form.

Bringing a factor under the radical

The same formula read from right to left puts a coefficient under the radical sign. Raise the coefficient to the power of the index and write it as a factor of the radicand.

• ${3\sqrt2=\sqrt{3^2\cdot 2}=\sqrt{18}}$
• ${2\sqrt[3]{5}=\sqrt[3]{2^3\cdot 5}=\sqrt[3]{40}}$
• ${a\sqrt[3]{x-b}=\sqrt[3]{a^3(x-b)}}$
• ${2ab\sqrt[3]{2ab^2}}={\sqrt[3]{8a^3b^3\cdot 2ab^2}}={\sqrt[3]{16a^4b^5}}$

In this way any positive number can be written as a radical with any index: ${3=\sqrt9=\sqrt[3]{27}=\sqrt[4]{81}},$ ${3a=\sqrt[4]{81a^4}}$ and ${a^2=\sqrt[3]{a^6}}.$

A negative coefficient stays in front of an even root. ${-3\sqrt2=-\sqrt{18}}.$ Squaring the $-3$ would lose the minus sign and give the positive number $\sqrt{18}.$ The same is true for a letter: if ${a\lt 0},$ then ${a\sqrt x=-\sqrt{a^2x}}.$

Changing the index. A common index

By the formula for changing the index, the index of a root and the exponent of its radicand may be multiplied or divided by the same number:

• ${\sqrt[6]{a^3}=\sqrt a}$
• ${\sqrt[4]{9}=\sqrt[4]{3^2}=\sqrt3}$
• ${\sqrt[6]{8}=\sqrt[6]{2^3}=\sqrt2}$
• ${\sqrt[6]{a^4}=\sqrt[3]{a^2}}$

This is the same as reducing the fraction in the exponent: ${a^{\frac36}=a^{\frac12}}.$

Only when the radicand is not negative. ${\sqrt[3]{-8}=-2},$ but ${\sqrt[6]{(-8)^2}=\sqrt[6]{64}=2}.$ With a negative radicand, take the minus sign out first: ${\sqrt[3]{-8}=-\sqrt[3]{8}=-\sqrt[6]{8^2}}.$

Radicals with different indices can be given a common index. Use the least common multiple of the indices, exactly as you find a common denominator for fractions.

• $\sqrt a$ and $\sqrt[3]{b}$: the least common multiple of 2 and 3 is 6, so ${\sqrt a=\sqrt[6]{a^3}}$ and ${\sqrt[3]{b}=\sqrt[6]{b^2}}.$
• $\sqrt2$ and $\sqrt[3]{3}$ become ${\sqrt[6]{2^3}=\sqrt[6]{8}}$ and ${\sqrt[6]{3^2}=\sqrt[6]{9}}.$
• $\sqrt[4]{a}$ and $\sqrt[6]{b}$ become $\sqrt[12]{a^3}$ and $\sqrt[12]{b^2}.$
• $\sqrt[3]{4}$ and $\sqrt[4]{5}$ become ${\sqrt[12]{4^4}=\sqrt[12]{256}}$ and ${\sqrt[12]{5^3}=\sqrt[12]{125}}.$

The root of a perfect square

When the radicand is the square of a sum or of a difference, the root can be taken exactly. Here the letters may be any real numbers, positive or negative, so the absolute value is needed. Recall the identities ${(x+y)^2=x^2+2xy+y^2}$ and ${(x-y)^2=x^2-2xy+y^2}.$

• ${\sqrt{x^2+2x+1}}={\sqrt{(x+1)^2}}={|x+1|}$
• ${\sqrt{a^2+a+\frac14}}={\sqrt{\left(a+\frac12\right)^2}}={\left|a+\frac12\right|}$
• ${\sqrt{x^2-6bx+9b^2}}={\sqrt{(x-3b)^2}}={|x-3b|}$

For example, ${x+1}$ may be negative. For ${x=-4}$ the radicand is ${16-8+1=9}$ and the root is 3, which is $|{-3}|.$

The same idea removes a root that stands inside another root. Look for two numbers whose sum is the term without a root and whose product is the number under the inner root, after that root is written with the coefficient 2. For ${3+2\sqrt2}$ the numbers are 1 and 2. For ${7-4\sqrt3=7-2\sqrt{12}}$ they are 4 and 3.

${\sqrt{3+2\sqrt2}}={\sqrt{1+2\sqrt2+2}}={\sqrt{\left(1+\sqrt2\right)^2}}={1+\sqrt2}$

${\sqrt{7-4\sqrt3}}={\sqrt{4-4\sqrt3+3}}={\sqrt{\left(2-\sqrt3\right)^2}}={\left|2-\sqrt3\right|}={2-\sqrt3}$

In the last step ${2-\sqrt3}$ is positive, because ${2=\sqrt4\gt\sqrt3}.$ A square root is never negative, so the order matters: ${\sqrt{\left(1-\sqrt2\right)^2}=\sqrt2-1}.$

Adding and Subtracting Radicals

Radicals with the same index and the same radicand are called like radicals. For example, $2\sqrt3$ and $5\sqrt3$ are like radicals, and so are $\sqrt[3]{x}$ and $-4\sqrt[3]{x}.$ The radicals $\sqrt3$ and $\sqrt5$ are not like radicals (different radicands), and neither are $\sqrt a$ and $\sqrt[3]{a}$ (different indices).

Twice the root of $a$ and three times the root of $a$ make five times the root of $a,$ just as ${2x+3x=5x}$:

${2\sqrt a+3\sqrt a=5\sqrt a}$

• ${2\sqrt3+5\sqrt3=7\sqrt3}$
• ${7\sqrt[3]{2}-4\sqrt[3]{2}=3\sqrt[3]{2}}$
• ${\sqrt5+\sqrt5=2\sqrt5}$
• ${\sqrt{ay}-3\sqrt{ay}=-2\sqrt{ay}}$
• ${3\sqrt[3]{h}-\left(-5\sqrt[3]{h}\right)=8\sqrt[3]{h}}$
• ${a\sqrt{b-h}+y\sqrt{b-h}}={(a+y)\sqrt{b-h}}$
• ${5\sqrt2+3\sqrt7-2\sqrt2+\sqrt7}={3\sqrt2+4\sqrt7}$

Simplify first

Radicals that look different may turn out to be like radicals after they are simplified.

• ${\sqrt8+\sqrt{50}}={2\sqrt2+5\sqrt2}={7\sqrt2}$
• ${\sqrt{50}-\sqrt8}={5\sqrt2-2\sqrt2}={3\sqrt2}$
• ${\sqrt{12}+\sqrt{27}-\sqrt{48}}={2\sqrt3+3\sqrt3-4\sqrt3}={\sqrt3}$
• ${\sqrt[3]{16}+\sqrt[3]{54}}={2\sqrt[3]{2}+3\sqrt[3]{2}}={5\sqrt[3]{2}}$
• ${\sqrt{16b}+\sqrt{4b}}={4\sqrt b+2\sqrt b}={6\sqrt b}$
• ${\sqrt{a^2x}+\sqrt{b^4x}}={a\sqrt x+b^2\sqrt x}={\left(a+b^2\right)\sqrt x}$
• ${\sqrt[3]{b^4y}-\sqrt[3]{by^4}}={b\sqrt[3]{by}-y\sqrt[3]{by}}={(b-y)\sqrt[3]{by}}$

Unlike radicals cannot be combined. The sum ${3\sqrt b+2\sqrt a}$ stays as it is: it is neither $5\sqrt a$ nor $5\sqrt b$ nor $5\sqrt{a+b}.$ The sum ${\sqrt a+\sqrt[3]{a}}$ cannot be simplified either. In particular, ${\sqrt2+\sqrt3\ne\sqrt5}$: the left side is about ${1.414+1.732=3.146},$ and ${\sqrt5\approx 2.236}.$

Multiplying Radicals

• ${\sqrt3\cdot\sqrt5=\sqrt{15}}$
• ${\sqrt2\cdot\sqrt{18}=\sqrt{36}=6}$
• ${\sqrt[3]{3}\cdot\sqrt[3]{9}=\sqrt[3]{27}=3}$
• ${\sqrt{8xb}\cdot\sqrt{2xb}}={\sqrt{16x^2b^2}}={4xb}$
• ${\sqrt{a+m}\cdot\sqrt{a-m}}={\sqrt{a^2-m^2}}$
• ${5\sqrt5\cdot 3\sqrt8}={15\sqrt{40}}={15\cdot 2\sqrt{10}}={30\sqrt{10}}$
• ${a\sqrt x\cdot b\sqrt x=ab\sqrt{x^2}=abx}$
• ${x\sqrt[3]{3}\cdot y\sqrt[3]{9}}={xy\sqrt[3]{27}}={3xy}$

As the second and third examples show, the product of two irrational numbers can be a whole number. In particular, ${\sqrt a\cdot\sqrt a=a}$: for example, ${\sqrt7\cdot\sqrt7=7}.$

Radicals with different indices

First rewrite the radicals with a common index, then multiply the radicands.

• ${\sqrt x\cdot\sqrt[3]{y}}={\sqrt[6]{x^3}\cdot\sqrt[6]{y^2}}={\sqrt[6]{x^3y^2}}$
• ${2\sqrt3\cdot 3\sqrt[3]{4}}={6\sqrt[6]{3^3}\cdot\sqrt[6]{4^2}}={6\sqrt[6]{27\cdot 16}}={6\sqrt[6]{432}}$
• ${\sqrt d\cdot\sqrt[3]{ab}}={\sqrt[6]{d^3}\cdot\sqrt[6]{a^2b^2}}={\sqrt[6]{a^2b^2d^3}}$

Roots of the same number are easiest to multiply as powers: add the exponents.

• ${\sqrt a\cdot\sqrt[3]{a}}={a^{\frac12}\cdot a^{\frac13}}={a^{\frac56}}={\sqrt[6]{a^5}}$
• ${\sqrt2\cdot\sqrt[4]{2}}={2^{\frac12+\frac14}}={2^{\frac34}}={\sqrt[4]{8}}$
• ${x\cdot\sqrt[n]{x}=x^{1+\frac1n}=x^{\frac{n+1}{n}}}$

Sums with radicals. Conjugates

Parentheses are removed in the usual way: every term is multiplied by every term.

${\sqrt2\left(\sqrt6+\sqrt8\right)}={\sqrt{12}+\sqrt{16}}={2\sqrt3+4}$

${\left(2+\sqrt3\right)\left(5-\sqrt3\right)}={10-2\sqrt3+5\sqrt3-3}={7+3\sqrt3}$

The identities for the square of a sum and of a difference work too:

${\left(a+\sqrt y\right)^2=a^2+2a\sqrt y+y}$

${\left(a-\sqrt y\right)^2=a^2-2a\sqrt y+y}$

${\left(\sqrt5+\sqrt2\right)^2}={5+2\sqrt{10}+2}={7+2\sqrt{10}}$

The most useful identity is the difference of squares. The expressions ${\sqrt a+\sqrt b}$ and ${\sqrt a-\sqrt b}$ are called conjugates, and so are ${a+\sqrt b}$ and ${a-\sqrt b}.$ The product of two conjugates has no radicals:

• ${\left(1+\sqrt2\right)\left(1-\sqrt2\right)}={1-2}={-1}$
• ${\left(3-2\sqrt2\right)\left(3+2\sqrt2\right)}={9-8}={1}$
• ${\left(\sqrt7+\sqrt3\right)\left(\sqrt7-\sqrt3\right)}={7-3}={4}$

Dividing Radicals

• ${\frac{\sqrt{50}}{\sqrt2}=\sqrt{25}=5}$
• ${\frac{\sqrt{6a^3x}}{\sqrt{3x}}=\sqrt{2a^3}=a\sqrt{2a}}$
• ${\frac{10\sqrt{27}}{2\sqrt3}=5\sqrt9=15}$
• ${\frac{10\sqrt[3]{108}}{5\sqrt[3]{4}}=2\sqrt[3]{27}=6}$
• ${\frac{8\sqrt{108}}{2\sqrt6}=4\sqrt{18}=12\sqrt2}$
• ${\frac{24x\sqrt{ay}}{6\sqrt a}=4x\sqrt y}$

Radicals with different indices are first rewritten with a common index:

• ${\frac{\sqrt[3]{a}}{\sqrt b}=\frac{\sqrt[6]{a^2}}{\sqrt[6]{b^3}}=\sqrt[6]{\frac{a^2}{b^3}}}$
• ${\frac{2\sqrt[3]{bc}}{3\sqrt{ac}}}={\frac23\sqrt[6]{\frac{b^2c^2}{a^3c^3}}}={\frac23\sqrt[6]{\frac{b^2}{a^3c}}}$
• ${\frac{\sqrt[6]{x^3y^2}}{\sqrt[3]{y}}}={\frac{\sqrt[6]{x^3y^2}}{\sqrt[6]{y^2}}}={\sqrt[6]{x^3}}={\sqrt x}$

Roots of the same number are easiest to divide as powers: subtract the exponents.

• ${\frac{\sqrt a}{\sqrt[6]{a}}=a^{\frac12-\frac16}=a^{\frac13}=\sqrt[3]{a}}$
• ${\frac{a^2}{\sqrt[3]{a}}=a^{2-\frac13}=a^{\frac53}=a\sqrt[3]{a^2}}$

Rationalizing the denominator

A fraction is easier to work with when there are no radicals in its denominator. For example, ${\frac{1}{\sqrt2}=\frac{\sqrt2}{2}},$ and dividing 1.414… by 2 is much easier than dividing 1 by 1.414… Removing the radicals from a denominator is called rationalizing the denominator. We multiply the numerator and the denominator by the same number, so the value of the fraction does not change.

1. The denominator is a square root. Multiply the numerator and the denominator by that root.

${\frac{2}{\sqrt3}=\frac{2\cdot\sqrt3}{\sqrt3\cdot\sqrt3}=\frac{2\sqrt3}{3}}$

${\frac{6}{\sqrt2}=\frac{6\sqrt2}{2}=3\sqrt2}$

${\frac{\sqrt a}{\sqrt x}=\frac{\sqrt a\cdot\sqrt x}{\sqrt x\cdot\sqrt x}=\frac{\sqrt{ax}}{x}}$

${\sqrt{\frac37}}={\frac{\sqrt3}{\sqrt7}}={\frac{\sqrt3\cdot\sqrt7}{\sqrt7\cdot\sqrt7}}={\frac{\sqrt{21}}{7}}$

2. The denominator is a root with a higher index. Multiply by the root that turns the radicand into a perfect $n$-th power, because ${\sqrt[n]{x}\cdot\sqrt[n]{x^{n-1}}=\sqrt[n]{x^n}=x}.$

${\frac{1}{\sqrt[3]{2}}=\frac{\sqrt[3]{2^2}}{\sqrt[3]{2}\cdot\sqrt[3]{2^2}}=\frac{\sqrt[3]{4}}{2}}$

${\frac{6}{\sqrt[4]{5}}=\frac{6\sqrt[4]{5^3}}{\sqrt[4]{5}\cdot\sqrt[4]{5^3}}=\frac{6\sqrt[4]{125}}{5}}$

3. The denominator is a sum or a difference that contains square roots. Multiply by the conjugate. The denominator becomes a difference of squares, and the roots disappear.

${\frac{\sqrt2}{3-\sqrt2}}={\frac{\sqrt2\left(3+\sqrt2\right)}{\left(3-\sqrt2\right)\left(3+\sqrt2\right)}}={\frac{3\sqrt2+2}{9-2}}={\frac{2+3\sqrt2}{7}}$

${\frac{4}{\sqrt7-\sqrt3}}={\frac{4\left(\sqrt7+\sqrt3\right)}{7-3}}={\sqrt7+\sqrt3}$

${\frac{a-\sqrt b}{a+\sqrt b}}={\frac{\left(a-\sqrt b\right)^2}{\left(a+\sqrt b\right)\left(a-\sqrt b\right)}}={\frac{a^2-2a\sqrt b+b}{a^2-b}}$   (for ${a^2\ne b}$)

A harder example. A denominator with three roots needs two steps. For the fraction $\frac{1}{\sqrt{10}-\sqrt2-\sqrt3},$ first group the last two roots and multiply by ${\sqrt{10}+\left(\sqrt2+\sqrt3\right)}$:

${\left(\sqrt{10}-\sqrt2-\sqrt3\right)}\cdot {\left(\sqrt{10}+\sqrt2+\sqrt3\right)}={10-\left(\sqrt2+\sqrt3\right)^2}={5-2\sqrt6}$

Then multiply by the conjugate ${5+2\sqrt6},$ because ${\left(5-2\sqrt6\right)\left(5+2\sqrt6\right)}={25-24}={1}.$ So

${\frac{1}{\sqrt{10}-\sqrt2-\sqrt3}}={\left(\sqrt{10}+\sqrt2+\sqrt3\right)\left(5+2\sqrt6\right)}$

Sometimes the opposite is needed, and the numerator is made rational in the same way: ${\frac{\sqrt a}{\sqrt x}=\frac{\sqrt a\cdot\sqrt a}{\sqrt x\cdot\sqrt a}=\frac{a}{\sqrt{ax}}}.$

Powers and Roots of Radicals

Powers

• ${\left(\sqrt[3]{2}\right)^2=\sqrt[3]{2^2}=\sqrt[3]{4}}$
• ${\left(\sqrt3\right)^4=\sqrt{3^4}=3^2=9}$
• ${\left(2\sqrt3\right)^2=2^2\cdot 3=12}$
• ${\left(2\sqrt5\right)^3}={8\sqrt{5^3}}={8\cdot 5\sqrt5}={40\sqrt5}$
• ${\left(\sqrt[3]{b+x}\right)^3=b+x}$
• ${\left(3a\sqrt[3]{y}\right)^3=27a^3y}$
• ${\left(a\sqrt[n]{x}\right)^2=a^2\sqrt[n]{x^2}}$
• ${\left(\sqrt[6]{a}\right)^2=\sqrt[6]{a^2}=\sqrt[3]{a}}$

Roots

• ${\sqrt{\sqrt[3]{a}}=\sqrt[6]{a}}$
• ${\sqrt[3]{\sqrt{64}}=\sqrt[6]{64}=2}$
• ${\sqrt{\sqrt x}=\sqrt[4]{x}}$ and ${\sqrt{\sqrt{\sqrt x}}=\sqrt[8]{x}}$
• ${\sqrt[3]{a\sqrt{xy}}=\sqrt[3]{a}\cdot\sqrt[6]{xy}}$

When there is a coefficient in front of the inner radical, bring it under that radical first:

${\sqrt{2\sqrt2}}={\sqrt{\sqrt{2^2\cdot 2}}}={\sqrt{\sqrt8}}={\sqrt[4]{8}}$

Fractional exponents do the same work without any radical signs:

${\sqrt{a\sqrt[3]{a}}}={\left(a\cdot a^{\frac13}\right)^{\frac12}}={\left(a^{\frac43}\right)^{\frac12}}={a^{\frac23}}={\sqrt[3]{a^2}}$

Comparing Radicals

• Same index. ${\sqrt7\lt\sqrt{10}}$ and ${\sqrt[3]{5}\gt\sqrt[3]{4}}.$
• A number and a radical. Which is larger, 4 or $\sqrt{15}?$ Since ${4=\sqrt{16}},$ we get ${4\gt\sqrt{15}}.$
• Coefficients. Which is larger, $3\sqrt2$ or $2\sqrt5?$ Bring the coefficients under the radicals: ${3\sqrt2=\sqrt{18}}$ and ${2\sqrt5=\sqrt{20}}.$ So ${3\sqrt2\lt 2\sqrt5}.$
• Different indices. Which is larger, $\sqrt2$ or $\sqrt[3]{3}?$ Use the common index 6: ${\sqrt2=\sqrt[6]{8}}$ and ${\sqrt[3]{3}=\sqrt[6]{9}}.$ So ${\sqrt2\lt\sqrt[3]{3}}.$
• Estimating. ${16\lt 20\lt 25},$ so ${4\lt\sqrt{20}\lt 5}.$ In the same way ${27\lt 50\lt 64},$ so ${3\lt\sqrt[3]{50}\lt 4}.$

For a number greater than 1, a root is smaller than the number, and the higher the index, the smaller the root: ${\sqrt{64}=8},$ ${\sqrt[3]{64}=4},$ ${\sqrt[6]{64}=2}.$ For a number between 0 and 1 it is the other way around, and a root is larger than the number: ${\sqrt{0.25}=0.5}.$

Graphs of Root Functions

Graph of the square root

The function ${y=\sqrt x}$ is defined only for ${x\ge 0},$ and its values are not negative: ${y\ge 0}.$ The graph starts at the point ${(0,\ 0)}$ and rises all the time, so the function is increasing. But it rises more and more slowly: to climb from 0 to 1 it needs 1 unit of $x,$ from 1 to 2 it needs 3 more units, and from 2 to 3 it needs 5 more.

The graph of ${y=\sqrt x}$ passes through ${(0,\ 0)},$ ${(1,\ 1)},$ ${(4,\ 2)}$ and ${(9,\ 3)}.$

The graph is half of a parabola lying on its side. The function ${y=\sqrt x}$ is the inverse of ${y=x^2}$ for ${x\ge 0}$: if ${y=x^2},$ then ${x=\sqrt y}.$ That is why the two graphs are mirror images of each other in the line ${y=x}.$

${y=x^2}$ for ${x\ge 0}$ and ${y=\sqrt x}$ are mirror images of each other in the line ${y=x}.$

Graph of the cube root

The function ${y=\sqrt[3]{x}}$ is defined for every real number $x,$ and it takes every real value. It is increasing everywhere, and it is the inverse of ${y=x^3}.$ Since ${\sqrt[3]{-x}=-\sqrt[3]{x}},$ the graph is symmetric about the origin.

The graph of ${y=\sqrt[3]{x}}$ passes through ${(-8,\ -2)},$ ${(-1,\ -1)},$ ${(0,\ 0)},$ ${(1,\ 1)}$ and ${(8,\ 2)}.$

The graphs of all roots with an even index look like the graph of $\sqrt x,$ and those with an odd index look like the graph of $\sqrt[3]{x}.$ Every graph ${y=\sqrt[n]{x}}$ passes through the points ${(0,\ 0)}$ and ${(1,\ 1)}.$

Practice Problems

Try them yourself, then open the answer. All letters stand for positive numbers, except in problem 15, where $x$ is any real number.

1. Evaluate: a) $\sqrt{121};$ b) $\sqrt[3]{-64};$ c) $\sqrt[4]{81};$ d) $\sqrt[5]{\frac{1}{32}}.$

a) 11, because ${11^2=121};$ b) $-4,$ because ${(-4)^3=-64};$ c) 3, because ${3^4=81};$ d) $\frac12,$ because ${\left(\frac12\right)^5=\frac{1}{32}}.$

2. Evaluate: a) $27^{\frac23};$ b) $16^{-\frac34};$ c) $0.04^{0.5};$ d) $32^{\frac35}.$

a) ${\left(\sqrt[3]{27}\right)^2=3^2=9};$ b) ${\frac{1}{\left(\sqrt[4]{16}\right)^3}=\frac{1}{2^3}=\frac18};$ c) ${\sqrt{0.04}=0.2};$ d) ${\left(\sqrt[5]{32}\right)^3=2^3=8}.$

3. Between which two consecutive whole numbers does each root lie? a) $\sqrt{75};$ b) $\sqrt[3]{100}.$

a) ${64\lt 75\lt 81},$ so ${8\lt\sqrt{75}\lt 9}.$ b) ${64\lt 100\lt 125},$ so ${4\lt\sqrt[3]{100}\lt 5}.$

4. Write with a radical sign: a) $a^{\frac35};$ b) $x^{-\frac12}.$ Write as a power: c) $\sqrt[4]{b^3};$ d) $\frac{1}{\sqrt[3]{y^2}};$ e) $\sqrt[6]{(a+b)^{-3}}.$

a) $\sqrt[5]{a^3};$ b) $\frac{1}{\sqrt x};$ c) $b^{\frac34};$ d) $y^{-\frac23};$ e) ${(a+b)^{-\frac36}=(a+b)^{-\frac12}}.$

5. Simplify: a) $\sqrt{72};$ b) $\sqrt{300};$ c) $\sqrt[3]{250};$ d) $\sqrt[4]{162}.$

a) ${\sqrt{36\cdot 2}=6\sqrt2};$ b) ${\sqrt{100\cdot 3}=10\sqrt3};$ c) ${\sqrt[3]{125\cdot 2}=5\sqrt[3]{2}};$ d) ${\sqrt[4]{81\cdot 2}=3\sqrt[4]{2}}.$

6. Simplify: a) $\sqrt{75a^4b^3};$ b) $\sqrt[3]{-125a^3x^6};$ c) $\sqrt{\frac{4a^4}{9x^2y^2}};$ d) $\sqrt[5]{\frac{32a^5x^{10}}{243}}.$

a) ${\sqrt{25a^4b^2\cdot 3b}=5a^2b\sqrt{3b}};$ b) $-5ax^2;$ c) $\frac{2a^2}{3xy};$ d) $\frac{2ax^2}{3}.$

7. Bring the coefficient under the radical: a) $5\sqrt3;$ b) $2\sqrt[3]{7};$ c) $-2\sqrt6.$ d) Write $ax^2$ as a sixth root.

a) ${\sqrt{25\cdot 3}=\sqrt{75}};$ b) ${\sqrt[3]{8\cdot 7}=\sqrt[3]{56}};$ c) ${-\sqrt{4\cdot 6}=-\sqrt{24}};$ d) $\sqrt[6]{a^6x^{12}}.$

8. Add or subtract: a) ${\sqrt{18a}+3\sqrt{2a}};$ b) ${3\sqrt{20}-\sqrt{45}+\sqrt{80}};$ c) ${\sqrt[3]{81}-\sqrt[3]{24}};$ d) ${\sqrt{12}+\sqrt{27}-\sqrt{75}}.$

a) ${3\sqrt{2a}+3\sqrt{2a}=6\sqrt{2a}};$ b) ${6\sqrt5-3\sqrt5+4\sqrt5=7\sqrt5};$ c) ${3\sqrt[3]{3}-2\sqrt[3]{3}=\sqrt[3]{3}};$ d) ${2\sqrt3+3\sqrt3-5\sqrt3=0}.$

9. Multiply: a) ${\sqrt6\cdot\sqrt{15}};$ b) ${2\sqrt3\cdot 5\sqrt{12}};$ c) ${\sqrt[3]{4}\cdot\sqrt[3]{10}};$ d) ${\sqrt2\cdot\sqrt[3]{4}}.$

a) ${\sqrt{90}=3\sqrt{10}};$ b) ${10\sqrt{36}=60};$ c) ${\sqrt[3]{40}=2\sqrt[3]{5}};$ d) ${\sqrt[6]{2^3}\cdot\sqrt[6]{4^2}=\sqrt[6]{128}=2\sqrt[6]{2}}.$

10. Remove the parentheses: a) ${\left(3+\sqrt5\right)\left(3-\sqrt5\right)};$ b) ${\left(\sqrt7-2\right)^2};$ c) ${\left(\sqrt3+\sqrt2\right)\left(2\sqrt3-\sqrt2\right)};$ d) ${\left(a-\sqrt b\right)^3}.$

a) ${9-5=4};$ b) ${7-4\sqrt7+4=11-4\sqrt7};$ c) ${6-\sqrt6+2\sqrt6-2=4+\sqrt6};$ d) ${a^3-3a^2\sqrt b+3ab-b\sqrt b}.$

11. Divide: a) $\frac{\sqrt{98}}{\sqrt2};$ b) $\frac{12\sqrt{30}}{4\sqrt6};$ c) $\frac{\sqrt[3]{54}}{\sqrt[3]{2}};$ d) $\frac{\sqrt[6]{a^2b^2d^3}}{\sqrt d}.$

a) ${\sqrt{49}=7};$ b) $3\sqrt5;$ c) ${\sqrt[3]{27}=3};$ d) ${\frac{\sqrt[6]{a^2b^2d^3}}{\sqrt[6]{d^3}}=\sqrt[6]{a^2b^2}=\sqrt[3]{ab}}.$

12. Rationalize the denominator: a) $\frac{10}{\sqrt5};$ b) $\frac{3}{\sqrt[3]{9}};$ c) $\frac{1}{\sqrt5+\sqrt3};$ d) $\frac{1+\sqrt2}{1-\sqrt2}.$

a) ${\frac{10\sqrt5}{5}=2\sqrt5};$ b) ${\frac{3\sqrt[3]{3}}{\sqrt[3]{9}\cdot\sqrt[3]{3}}=\frac{3\sqrt[3]{3}}{3}=\sqrt[3]{3}};$ c) ${\frac{\sqrt5-\sqrt3}{5-3}=\frac{\sqrt5-\sqrt3}{2}};$ d) ${\frac{\left(1+\sqrt2\right)^2}{1-2}=-3-2\sqrt2}.$

13. Simplify: a) ${\left(2\sqrt3\right)^4};$ b) $\sqrt{\sqrt[3]{x^{12}}};$ c) $\sqrt[3]{\sqrt{27}};$ d) $\sqrt{x\sqrt x}.$

a) ${16\cdot 9=144};$ b) ${\sqrt[6]{x^{12}}=x^2};$ c) ${\sqrt[6]{3^3}=\sqrt3};$ d) ${\sqrt{\sqrt{x^3}}=\sqrt[4]{x^3}}.$

14. Which number is larger? a) $2\sqrt7$ or $3\sqrt3;$ b) $\sqrt[3]{3}$ or $\sqrt2;$ c) 5 or $\sqrt[3]{120}.$

a) ${2\sqrt7=\sqrt{28}}$ and ${3\sqrt3=\sqrt{27}},$ so $2\sqrt7$ is larger. b) ${\sqrt[3]{3}=\sqrt[6]{9}}$ and ${\sqrt2=\sqrt[6]{8}},$ so $\sqrt[3]{3}$ is larger. c) ${5=\sqrt[3]{125}},$ so 5 is larger.

15. Simplify: a) $\sqrt{x^2-10x+25};$ b) $\sqrt{11-6\sqrt2}.$

a) ${\sqrt{(x-5)^2}=|x-5|};$ b) ${\sqrt{9-6\sqrt2+2}}={\sqrt{\left(3-\sqrt2\right)^2}}={3-\sqrt2}.$ The difference ${3-\sqrt2}$ is positive, because ${3\gt\sqrt2}.$

16. Solve the equations: a) ${x^2=48};$ b) ${x^3=-54};$ c) ${x^4=81};$ d) ${x^2=-9}.$

a) ${x=\pm\sqrt{48}=\pm 4\sqrt3};$ b) ${x=\sqrt[3]{-54}=-3\sqrt[3]{2}};$ c) ${x=\pm\sqrt[4]{81}=\pm 3};$ d) no real solutions, because a square is never negative.

17. A square has an area of 50 square centimeters. A cube has a volume of 54 cubic centimeters. Find the side of the square and the edge of the cube.

The side is ${\sqrt{50}=5\sqrt2\approx 7.07}$ centimeters. The edge is ${\sqrt[3]{54}=3\sqrt[3]{2}\approx 3.78}$ centimeters.

18. The legs of a right triangle are 2 and 4. Find the hypotenuse.

By the Pythagorean theorem, the hypotenuse is ${\sqrt{2^2+4^2}}={\sqrt{20}}={2\sqrt5}\approx{4.47}.$

19. Show that ${\sqrt{4+2\sqrt3}-\sqrt{4-2\sqrt3}=2}.$

${4+2\sqrt3=\left(\sqrt3+1\right)^2}$ and ${4-2\sqrt3=\left(\sqrt3-1\right)^2}.$ Both ${\sqrt3+1}$ and ${\sqrt3-1}$ are positive, so the difference is ${\left(\sqrt3+1\right)-\left(\sqrt3-1\right)=2}.$

20. Solve the equations: a) ${\sqrt x=5};$ b) ${\sqrt x=-5};$ c) ${\sqrt[3]{x}=-2}.$

a) ${x=5^2=25};$ b) no solutions, because a square root is never negative; c) ${x=(-2)^3=-8}.$

See also: Exponents · Absolute Values